Step 1: Assume a 100 g sample of the rubber.
- Butadiene = 70 g, Isoprene = 20 g, Sulfur = 5 g, Carbon black = 5 g.
Step 2: Moles of each monomer.
\[
n_{\text{butadiene}} = \frac{70}{54} = 1.296 \; \text{mol}
\]
\[
n_{\text{isoprene}} = \frac{20}{68} = 0.294 \; \text{mol}
\]
Total moles of monomer units:
\[
n_{\text{total}} = 1.296 + 0.294 = 1.59 \; \text{mol}
\]
Step 3: Moles of sulfur atoms.
\[
n_S = \frac{5}{32} = 0.156 \; \text{mol of sulfur atoms}
\]
Each crosslink consumes on average 2 sulfur atoms →
\[
n_{\text{crosslinks}} = \frac{0.156}{2} = 0.078 \; \text{mol crosslinks}
\]
Step 4: Maximum possible crosslink sites.
- Each monomer unit (butadiene or isoprene) contains one double bond → potential site for crosslinking.
- So maximum possible crosslinks ≈ total number of monomer units = 1.59 mol.
Step 5: Fraction of crosslinked sites.
\[
% \text{ crosslinks} = \frac{n_{\text{crosslinks}}}{n_{\text{total}}} \times 100
\]
\[
= \frac{0.078}{1.59} \times 100 = 4.9%
\]
Check carefully: Some formulations take crosslink as connecting two monomers, i.e. each crosslink consumes 2 monomer sites. So maximum crosslinks = \(n_{\text{total}}/2 = 0.795 \; mol\).
Thus,
\[
% \text{ crosslinks} = \frac{0.078}{0.795} \times 100 = 9.8%
\]
Since question states “percentage of possible crosslinks” (meaning per two sites), the correct interpretation is second case.
Final Answer:
\[
\boxed{9.8%}
\]